M216: Exercise sheet 1

1. Let \(U\) be a subset of a vector space \(V\). Show that \(U\) is a linear subspace of \(V\) if and only if \(U\) satisfies the following conditions:

(i) \(0\in U\);

(ii) For all \(u_1,u_2\in U\) and \(\lambda \in \F \), \(u_1+\lambda u_2\in U\).


2. Which of the following subsets of \(\R ^3\) are linear subspaces? In each case, briefly justify your answer.
(a) \(U_1:=\set {(x_1,x_2,x_3)\st x_1^2+x_2^2+x_3^2=1}\) (b) \(U_2:=\set {(x_1,x_2,x_3)\st x_1=x_2}\) (c) \(U_3:=\set {(x_1,x_2,x_3)\st x_1+2x_2+3x_3=0}\)

3. Which of the following maps \(f:\R ^2\to \R ^2\) are linear? In each case, briefly justify your answer.
(a) \(f(x,y)=(5x+y,3x2y)\) (b) \(f(x,y)=(5x+2,7y)\) (c) \(f(x,y)=(\cos y,\sin x)\) (d) \(f(x,y)=(3y^{2},x^3)\).
Homework

4. Let \(\cI \) be a set and \(V\) a vector space over a field \(\F \). Recall that \(V^{\cI }\) is the set of maps \(\cI \to V\).
Show that \(V^{\cI }\) is a vector space under pointwise addition and scalar multiplication.

5. Let \(\R [x]\) be the space of real polynomials. This is a vector space under coefficientwise addition and scalar multiplication.
For \(d\in \N \), let \(P_{d}\subset \R [x]\) be the set of polynomials of degree no more than \(d\). Show that \(P_d\leq \R [x]\) and has basis \(1,x,\dots , x^{d}\)
Define a linear map \(D:P_d\to P_d\) by \(D(p)=p'\). Compute its matrix with respect to \(1,x,\dots , x^{d}\). What are \(\ker D\) and \(\im D\)?
Additional questions

6. Which of the following subsets of \(\C ^3\) are linear subspaces over \(\C \)? In each case, briefly justify your answer.
(a) \(U_1:=\set {(z_1,z_2,z_3)\st z_1z_2=1}\) (b) \(U_2:=\set {(z_1,z_2,z_3)\st z_1=\bar {z}_2}\) (c) \(U_3:=\set {(z_1,z_2,z_3)\st z_1+\sqrt {1}z_2+3z_3=0}\)

7. Let \(V\) be an \(n\)dimensional vector space over \(\C \), and let \(V_\R \) be the underlying vector space over \(\R \) (thus \(V_\R \) has the same set of vectors as \(V\), but scalar multiplication is restricted to real scalars). Prove that \(V_\R \) has dimension \(2n\).
[Hint: let \(\cB :v_1,v_2,\ldots ,v_n\) be a basis for \(V\) and show that \(\cB _\R :v_1,i v_1, v_2,i v_2, \ldots ,v_n,i v_n\) is a basis for \(V_\R \), where \(i\in \C \) is \(\sqrt {1}\) rather than an index!]
Warmup questions
Please hand in at 4W level 1 by NOON on Friday 13th October 2023
M216: Exercise sheet 1—Solutions

1. First suppose that \(U\leq V\). The \(U\) is nonempty so there is some \(u\in U\) and then, since \(U\) is closed under addition and scalar multiplication, \(0=u+(1)u\in U\) also and condition (i) is satisfied. Now if \(u_1, u_2\in U\) and \(\lambda \in \F \), then \(\lambda u_2\in U\) (\(U\) is closed under scalar multiplication) and so \(u_1+\lambda u_2\in U\) (\(U\) is closed under addition). Thus condition (ii) holds also.
For the converse, if conditions (i) and (ii) hold, then, first, \(0\in U\) so \(U\) is nonempty and, second, \(U\) is closed under addition (take \(\lambda =1\) in condition (ii)) and under scalar multiplication (take \(u_1=0\) in condition (ii)). Thus \(U\leq V\).

2.

(a) \(U_1\) is not a subspace as it does not contain \(0\)!

(b) \(U_2\) is a subspace: in fact, it is \(\ker \phi _A\) where \(A= \begin {pmatrix} 1&1&0 \end {pmatrix} \).

(c) \(U_3\) is a subspace. It is \(\ker \phi _A\) for \(A= \begin {pmatrix} 1&2&3 \end {pmatrix} \).


3.

(a) Here \(f\) is linear: it is the map \(\phi _A\) corresponding to the matrix
\(\seteqnumber{0}{}{0}\)\begin{equation*} A= \begin{pmatrix} 5&1\\3&2 \end {pmatrix}. \end{equation*}

(b) This is not linear (because of that \(+2\) term). In particular \(f(0,0)=(2,0)\neq 0\)!

(c) Again \(f(0,0)=(1,0)\neq 0\) so this \(f\) cannot be linear. Of course, we already know this because it is certainly not true that \(\cos (y_1+y_2)=\cos y_1+\cos y_2\).

(d) Another nonlinear map: for example \(f(2x,2y)\neq 2f(x,y)\).


4. The basic idea is that the vector space axioms for \(V^{\cI }\) will follow from those of \(V\) applied to the values of elements of \(V^{\cI }\). Since those elements are completely determined by their values, this will bake the cake.
In more detail: let \(u,v,w\in V^{\cI }\), then, for \(i\in \cI \),
\(\seteqnumber{0}{}{0}\)\begin{equation*} (u+v)(i)=u(i)+v(i)=v(i)+u(i)=(v+u)(i), \end{equation*}
whence \(u+v=v+u\). Here the first and last equalities are just the definition of pointwise addition and the middle one of commutativity of addition in \(V\).
Similarly,
\(\seteqnumber{0}{}{0}\)\begin{equation*} ((u+v)+w)(i)=(u+v)(i)+w(i)=(u(i)+v(i))+w(i)= u(i)+(v(i)+w(i))=(u+(v+w))(i) \end{equation*}
so that \((u+v)+w=u+(v+w)\).
The zero element is the zero map defined by \(0(i):=0\), for all \(i\in \cI \), while the additive inverse \(v\) of \(v\in V^{\cI }\) is defined by \((v)(i):=(v(i))\). Now
\(\seteqnumber{0}{}{0}\)\begin{align*} (v+0)(i)&=v(i)+0(i)=v(i)+0=v(i)\\ (v+(v))(i)&=v(i)+(v)(i)=v(i)v(i)=0=0(i) \end{align*} so that \(v+0=v\) and \(v+(v)=0\) as required.
The axioms around scalar multiplication are verified in the same way. For example, for \(\lambda ,\mu \in \F \),
\(\seteqnumber{0}{}{0}\)\begin{equation*} ((\lambda +\mu )v)(i)=(\lambda +\mu )(v(i))=\lambda (v(i))+\mu (v(i))=(\lambda v)(i)+(\mu v)(i)=(\lambda v+\mu v)(i) \end{equation*}
so that \((\lambda +\mu )v=\lambda v+\mu v\).
Again, for \(u,v\in V^{\cI }\) and \(\lambda \in \F \),
\(\seteqnumber{0}{}{0}\)\begin{multline*} (\lambda (u+v))(i)=\lambda (u+v)(i)=\lambda (u(i)+v(i))=\lambda u(i)+\lambda v(i)\\ =(\lambda u)(i)+(\lambda v)(i)=(\lambda u+\lambda v)(i) \end{multline*} so that \(\lambda (u+v)=\lambda u+\lambda v\).
For \(\lambda ,\mu \in F\) and \(v\in V^{\cI }\),
\(\seteqnumber{0}{}{0}\)\begin{equation*} ((\lambda \mu )v)(i)=(\lambda \mu )v(i)=\lambda (\mu v(i))=(\lambda (\mu v))(i) \end{equation*}
so that \((\lambda \mu )v=\lambda (\mu v)\).
Finally, \((1v)(i)=1v(i)=v(i)\) so that \(1v=v\) and we are (at last!) done.

5. Clearly \(P_d\) is nonempty as it contains the zero polynomial. Moreover, for any polynomials \(p,q\) and \(\lambda \in \R \), we have
\(\seteqnumber{0}{}{0}\)\begin{align*} \deg (p+q)&\leq \max \set {\deg p,\deg q}\\ \deg (\lambda p)&\leq \deg p, \end{align*} from which it easily follows that \(P_d\) is closed under addition and scalar multiplication.
Any polynomial \(p\in P_d\) has a unique expression of the form
\(\seteqnumber{0}{}{0}\)\begin{equation*} p=a_0+a_1x+\dots +a_dx^{d}. \end{equation*}
It now follows from Proposition 1.1 that \(1,\rng {x}{x^d}\) is a basis for \(P_d\).
Set \(v_j=x^{j1}\), for \(1\leq j\leq d+1\), and compute \(D v_j\) in terms of the \(v_i\):
\(\seteqnumber{0}{}{0}\)\begin{equation*} D v_j=(j1)v_{j1} \end{equation*}
so that the matrix \(A\) of \(D\) with respect to this basis has all entries 0 except just above the diagonal where \(A_{(j1)j}=j1\). For example, if \(d=3\), we have
\(\seteqnumber{0}{}{0}\)\begin{equation*} A= \begin{pmatrix} 0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0 \end {pmatrix}. \end{equation*}
The kernel of \(D\) is the constant polynomials \(P_0\) and the image is \(P_{d1}\).

6.

(a) \(0\notin U_1\) so \(U_1\) is not a subspace.

(b) \(U_2\) is not a subspace because it is not closed under complex scalar multiplication: \((1,1,0)\in U_2\) but \(i(1,1,0)=(i,i,0)\) is not (here \(i=\sqrt {1}\)). In general, any time you see complex conjugation in the definition of a subset, it is unlikely to be a complex subspace.

(c) \(U_3=\ker \phi _A\) for \(A= \begin {pmatrix} 1&\sqrt {1}&3 \end {pmatrix} \) and so is a subspace.


7. Following the hint we need to show that any \(v\in V_\R \) can be written uniquely as a real linear combination of vectors in the list \(\cB _\R \). Since \(v\in V\), we may write \(v=\sum _{j=1}^n \lambda _j v_j\) for unique \(\lambda _j\in \C \). Write \(\lambda _j=a_j+i b_j\) with \(a_j,b_j\in \R \). Then \(v=\sum _{j=1}^n (a_j v_j+b_j i v_j)\) and this expression is unique: it suffices to observe that for \(v=0\), \(\lambda _j=0\) for all \(j\), and hence \(a_j=b_j=0\) for all \(j\).